Isn't this just asking for the number of equivalence classes under cyclic permutations? These are called necklaces (or possibly bracelets, I can never remember the difference) and are well studied. See oeis.org/A000031
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Gordon RoyleSep 14 '10 at 7:54

1

By definition, every cyclic conjugacy class of words of length n contains a unique maximal word. So this problem can be restated as: how many orbits are there for the action of the cyclic group of order n on the words of length n over alphabet {0,1} by cyclic rotations.
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Victor ProtsakSep 14 '10 at 7:54

@Gordon A necklaces is the equivalence class of a word under cyclic permutation, a bracelet is the equivalence class of a word under cyclic permutation and reflection.
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Mark BellSep 14 '10 at 17:32

2 Answers
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For aperiodic (sometimes also called, full period) strings, the term you are looking for is Lyndon words. These are the (unique lexicographically-least) representative of a full-period necklace (as stated in the comments, a necklace is the equivalence class under cyclic rotation). The number $k(n)$ you ask for is exactly the number of necklaces, and again, as stated in the comments, it is given by
$k(n)=\frac{1}{n}\sum_{d|n}\phi(d)2^{n/d}$. You can check out a proof for this in S.W.Golomb's book "Shift Register Sequences" (in the 1967 edition, start looking at around page 171 and look for the cycles of $PCR_n$).

Thanks to everybody! As a matter of fact, I had checked if the first few items of the sequence were already listed in the "Encyclopedia of integer sequences", but -alas!- I did not find the right answer because of a mistake in the count of 8-letters "maximal" words :-( I hope this will not affect my reputation ;-)
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ccarminatSep 16 '10 at 6:52